Let be the number such that
.
Then and (cf theorem 7.87),
so by
continuity of ,
and
. Since for all , it follows by the inequality theorem that
, and since , we have
. Hence,
.

9.13Exercise (Intermediate value theorem.)A
Let
with and let
be a continuous
function with . Let be a number in the interval
. Show that there is some with . (Use
theorem 9.11. Do not reprove it.)

9.14Notation ( is between and .)
Let
. I say is between and if either or
.

9.15Corollary (Intermediate value theorem.)
Let
with . Let
be a continuous function with
. If is any number between and , then there is some
such that . In particular, if and have opposite
signs, there is a number with .

Proof: By exercise 9.13A, the result holds when . If
, let . Then is continuous on and , so by
exercise 9.13A there is a with , so so
.

9.16Example.
Let be real numbers with , and let

We will show that there is a number
such that . Suppose, in order
to get a contradiction, that no such number exists, and let

(I use the fact that has no zeros here.) Then

It follows that for some , so and have opposite signs
for some , and is continuous on , so by the intermediate value
theorem, for some , contradicting the assumption that
is never
zero.

9.17Exercise.A
Give examples of the requested functions, or
explain why no such
function exists. Describe your functions by formulas if you can, but pictures of
graphs will do if a formula seems too complicated.

a)

, has no maximum.

b)

, is continuous, has no maximum.

c)

, is continuous, has no maximum or
minimum.

d)

, is bounded and continuous, has no maximum.

9.18Exercise.
Let . Prove that the equation has at least three
solutions in
.
s

9.20Note.
The intermediate value theorem was proved independently by
Bernhard Bolzano in
1817 [42], and Augustin Cauchy in 1821[23, pp 167-168]. The
proof we have given is almost identical with Cauchy's proof.